0
$\begingroup$

My physics teacher told me and some other students that when you experience negative g-forces, your weight equals: $$\frac{mass}{g_{force}}$$ So when the g-forces equal -4 g your mass will be: $$\frac{mass}{-4}$$ Now I don't think this is true, because this doesn't work when the g-forces are positive. And also to follow the pattern (example pattern):)

$45 \, \mathrm{kg} \times 3g = 1350\, \mathrm N$

$45 \, \mathrm{kg} \times 2g = 900\, \mathrm N$

$45 \, \mathrm{kg} \times 1g = 450 \, \mathrm N$

$45 \, \mathrm{kg} \times 0g = 0\, \mathrm N$

$45 \, \mathrm{kg} \times (-1g) = -450\, \mathrm N$

$45 \, \mathrm{kg}\ /\;(-1g) = -450 \, \mathrm N$

$45 \, \mathrm{kg} \times (-2g) = -900\, \mathrm N$

$45 \, \mathrm{kg}\ / \;(-2g) = -225\, \mathrm N$

Which one is true?

$\endgroup$
  • $\begingroup$ Either you have misunderstood your teacher, or the teacher made a mistake. You appear to be familiar with $F=ma$, and that equation is valid for positive & negative accelerations. $\endgroup$ – PM 2Ring Sep 21 at 20:42
  • $\begingroup$ @PM2Ring Yeha, to be honest my teacher told some friends who told me so there may have been some misinterpretation but idk. Thought it sounded weird anyway $\endgroup$ – Melvin Sep 21 at 21:15
  • $\begingroup$ @PM2Ring so in conclusion the patter with f=ma is correct? $\endgroup$ – Melvin Sep 21 at 21:17
1
$\begingroup$

The pattern with $F=ma$ is the correct one. The g-force experienced by an object can be written as $F = n(mg)$ where m is the mass; n is the number of g's and g is the gravitational constant. When n is greater than 1 the object will experience a force greater than the usual gravitational attraction of Earth; if we were to measure its weight with a scale when experiencing said force its apparent mass weight would be greater than if it were only under the effects of gravity (which corresponds to n=1). For $0 < n < 1$ an object would be under the influence of a force smaller than the one exerted by gravity; therefore, if we were to weight the object it would be lighter than usual. As you may have noticed the case of n=0 corresponds to the feeling of weightlessness which is what astronauts feel and corresponds to no apparent weight.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.